# Let T P_2 rarr RR^3 be a transformation given by T(f(x))=[(f(0)),(f(1)),(2f(1))] (a)Then show that T is a linear transformation. (b)Find and describe the kernel(null space) of T i.e Ker(T) and range of T. (c)Show that T is one-to-one.

Let T $$\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}$$ be a transformation given by
$$\displaystyle{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}$$
(a)Then show that T is a linear transformation.
(b)Find and describe the kernel(null space) of T i.e Ker(T) and range of T.
(c)Show that T is one-to-one.

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Elberte

The map T : $$\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}$$ be a transformation given by
$$\displaystyle{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}$$.
(a)Let f,$$\displaystyle{g}\in{P}_{{2}}$$ and k is a real number. Then
$$\displaystyle{T}{\left({f{{\left({x}\right)}}}+{k}{g{{\left({x}\right)}}}\right)}={T}{\left({f{{\left({f}+{k}{g}\right)}}}{\left({x}\right)}\right)}$$
$$\displaystyle={\left[\begin{array}{c} {\left({f}+{k}{g}\right)}{\left({0}\right)}\\{\left({f}+{k}{g}\right)}{\left({1}\right)}\\{2}{\left({f}+{k}{g}\right)}{f{{\left({1}\right)}}}\end{array}\right]}$$
$$\displaystyle={\left[\begin{array}{c} {f{{\left({0}\right)}}}+{k}{g{{\left({0}\right)}}}\\{\left({f{{\left({1}\right)}}}+{k}{g{{\left({1}\right)}}}\right)}\\{2}{f{{\left({1}\right)}}}+{2}{k}{g{{\left({1}\right)}}}\end{array}\right]}$$
$$\displaystyle={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}+{\left[\begin{array}{c} {k}{g{{\left({0}\right)}}}\\{k}{g{{\left({1}\right)}}}\\{2}{k}{g{{\left({1}\right)}}}\end{array}\right]}$$
$$=T(f(x))+kT(g(x)).$$
Therefore T:$$\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}$$ is a linear transformation.
(b)
$$\displaystyle{K}{e}{r}{\left({T}\right)}={\left\lbrace{f}\in{P}_{{2}}:{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {0}\\{0}\\{0}\end{array}\right]}\right\rbrace}$$
$$\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}={\left[\begin{array}{c} {0}\\{0}\\{0}\end{array}\right]}\right\rbrace}$$
$$\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{f{{\left({0}\right)}}}={0},{f{{\left({1}\right)}}}={0},{2}{f{{\left({1}\right)}}}={0}\right\rbrace}$$
$$\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{f{{\left({0}\right)}}}={0},{f{{\left({1}\right)}}}={0}\right\rbrace}$$
$$\displaystyle={\left\lbrace{f{{\left({x}\right)}}}={a}{x}{\left({x}-{1}\right)},{a}\in\mathbb{R}\right\rbrace}$$
Therefore Nullity $$(T)=1.$$
Again Range(T) $$\displaystyle={\left\lbrace{T}{\left({f{{\left({x}\right)}}}\right)}:{f}\in{P}_{{2}}\right\rbrace}={s}{p}{a}{n}{\left\lbrace{f{{\left({0}\right)}}},{f{{\left({1}\right)}}}\right\rbrace}$$. Therefore
Rank $$(T)=2$$.
(c)Since Nullity $$(T)=1$$ therefore T is not one-one.